Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Light of wavelength falling on a piece of metal ejects electrons with sufficient ener

Light of wavelength 330 \mathrm{~nm} falling on a piece of metal ejects electrons with sufficient energy which requires voltage V_0 to prevent them from reaching a collector. In the same setup, light of wavelength 220 \mathrm{~nm}, ejects electrons which require twice the voltage \mathrm{V}_0 to stop them in reaching a collector. The numerical value of voltage \mathrm{V}_0 is:
 

Option: 1

\frac{16}{15} \mathrm{~V}


Option: 2

\frac{15}{16} \mathrm{~V}


Option: 3

\frac{15}{8} \mathrm{~V}


Option: 4

\frac{8}{15} \mathrm{~V}


Answers (1)

best_answer

Let \mathrm{W} be the work function of metal. Then,
\begin{aligned} & \mathrm{eV}_0=\frac{\mathrm{hc}}{330 \times 10^{-9}}-\mathrm{W}\quad \quad \quad \quad \quad(i) \\ & \mathrm{e}\left(2 \mathrm{~V}_0\right)=\frac{\mathrm{hc}}{220 \times 10^{-9}}-\mathrm{W} \quad \quad \quad(ii) \end{aligned}

Solving these two equations, we get

\mathrm{V}_0=\frac{10^9 \times \mathrm{h} \times \mathrm{c}}{110 \times \mathrm{e} \times 6}=\frac{10^9 \times 6.6 \times 10^{-34} \times 3 \times 10^8}{110 \times 1.6 \times 10^{-19} \times 6}=\frac{15}{8} \mathrm{~V}
 

Posted by

qnaprep

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question